Animated Solution for Chemistry - Surface Chemistry: When silver nitrate solution is added to potassium iodide solution then the sol produced is

Preferential Adsorption

Have you ever wondered why tiny particles in a colloidal solution don't just clump together and settle at the bottom? The secret lies in a fascinating phenomenon called preferential adsorption, which creates an electrical shield around each particle. Let's dive into this classic problem and uncover the chemistry behind it.

The problem states: "When silver nitrate solution is added to potassium iodide solution..."

This phrasing is the key to the entire puzzle. In chemistry, when we say "A is added to B", it implies that B is the bulk medium sitting in the beaker. Therefore, in our scenario, potassium iodide ($\text{KI}$) is the dispersion medium and is present in excess.

When these two solutions mix, a double displacement reaction occurs:

The silver iodide ($\text{AgI}$) forms a solid precipitate. These tiny solid particles act as the core of our colloidal system.

Preferential Adsorption

Now, colloidal particles possess a very special property. They don't like to stay completely neutral in an ionic environment. Instead, they tend to adsorb ions onto their surface. But they are picky! A colloidal particle will preferentially adsorb an ion that is common to its own crystal lattice, provided that ion is present in excess in the surrounding solution.

Let's apply this rule to our setup. Our colloidal core is $\text{AgI}$, which is made of ${\text{Ag}}^{+}$ and ${\text{I}}^{-}$ ions. The surrounding solution has an excess of $\text{KI}$, which provides ${\text{K}}^{+}$ and ${\text{I}}^{-}$ ions.

What is the common ion? Exactly, the iodide ion (${\text{I}}^{-}$)!

Because ${\text{I}}^{-}$ is the common ion, the $\text{AgI}$ core will tightly adsorb iodide ions all over its surface. This creates what we call the primary or fixed layer.

Because this fixed layer is made entirely of negatively charged iodide ions, the entire colloidal particle acquires a net negative charge. We represent this specific colloidal sol as $\text{AgI}/{\text{I}}^{-}$.

To maintain overall electrical neutrality in the beaker, the positively charged potassium ions (${\text{K}}^{+}$) from the solution will be attracted to this negative fixed layer. They form a loose, mobile boundary around the particle known as the secondary or diffused layer. Together, the fixed layer and the diffused layer make up the Electrical Double Layer, which acts as a repulsive shield preventing the particles from coagulating.

Since the fixed layer determines the charge of the sol, our resulting sol is negatively charged and represented as $\text{AgI}/{\text{I}}^{-}$. This makes option (a) the correct answer.

A Quick Thought Experiment: What if the question had stated the reverse? What if potassium iodide was added to excess silver nitrate?

In that case, ${\text{AgNO}}_3$ would be in excess. The common ion between the $\text{AgI}$ core and the excess ${\text{AgNO}}_3$ would be the silver ion (${\text{Ag}}^{+}$). The core would adsorb ${\text{Ag}}^{+}$ ions, forming a positively charged sol represented as $\text{AgI}/{\text{Ag}}^{+}$.

Always pay close attention to which reagent is in excess—it completely flips the nature of the colloid!